Solution:
\( 131.76 = 10 \log \frac{I_T}{I_0} \)
\( 13.176 = \log(I_T) - \log(I_0) \)
\( 13.176 = \log(I_T) + 16 \)
\( -2.824 = \log (I_T) \)
\( I_T = 10 ^{-2.824} \)
\( 128.7 = 10 \log \frac{I_D}{I_0} \)
\( 12.87 = \log(I_D) - \log(I_0) \)
\( 12.87 = \log(I_D) + 16 \)
\( -3.13 = \log (I_D) \)
\( I_D = 10 ^{-3.13} \)
\( \frac{I_T}{I_0}= \frac{10^{-2.824}}{10^{-3.13}}=10^{0.306} \approx 2 \)
It was twice as intense.